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Question

# If any tangent plane to the sphere (xâˆ’a)2+(yâˆ’b)2+(zâˆ’c)2=r2 makes intercepts a,b,c with the coordinate axes at A,B,C and P be the center of the sphere. Then

A
1a2+1b2+1c2=4r2
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B
1a2+1b2+1c2=1r2
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C
Volume of the tetrahedron PABC=abc3
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D
Volume of the tetrahedron PABC=abc6
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Solution

## The correct options are A 1a2+1b2+1c2=4r2 C Volume of the tetrahedron PABC=abc3Equation of the plane is xa+yb+zc=1 Distance of the plane from the center of the sphere is 2√1a2+1b2+1c2=r1a2+1b2+1c2=4r2 Now in triangle ABC Distance of C from AB AB:x−aa=y−0−b=z−00=k any point on the line (a(k+1),−bk,0) Solving the point with the point C k=−a2a2+b2 So perpendicular distance will be =√a2b2a2+b2+c2=√a2b2+a2c2+b2c2a2+b2=2abcr√(a2+b2) AB=√a2+b2 ar.△ABC=122abcr√(a2+b2)⋅√a2+b2=abcr perpendicular distance between plane ABC and P =r ∴ volume =r3abcr=abc3

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